could anyone quickly help me find this radius...i am totally stuck here and have to know this before i can move forward.

Find the standard form of the equation -

center is (8, 3) and passes thru (-6, -1)

so far i have figured out (x-8)^2 + (y-3)^2 = r^2

the radius is what i cannot understand what to figure out...

2. Use the distance formula:

$d=\sqrt{(x-x_{1})^{2}+(y-y_{1})^{2}}$

3. Yes, you're on the right track. Now all you have to do is... Substitute $x = -6$ and $y = -1$ into the same equation to get $r$.

I hope that helps.

ILoveMaths07.

4. Originally Posted by shipwreck
could anyone quickly help me find this radius...i am totally stuck here and have to know this before i can move forward.

Find the standard form of the equation -

center is (8, 3) and passes thru (-6, -1)

so far i have figured out (x-8)^2 + (y-3)^2 = r^2

the radius is what i cannot understand what to figure out...
The radius of a circle is by definition, the distance from the center to any point on the circle. What is the distance from (8, 3) to (-6, -1)?

Another perfectly good way to do this is to say that, since (-6,-1) is ON the circle, it must satisfy that equation. If x=-6 and y= -1, what does your equation give you?